Chapter 11: Q. 50 (page 774)

Solve the system. Use any method you wish.
$\{\begin{cases} x^{3} - 2 x^{2} + y^{2} + 3 y - 4 = 0 \\ x - 2 + \frac{y^{2} - y}{x^{2}} = 0 \end{cases}$

Short Answer

Expert verified

The solution of system of equations are $(2, 1)$ .

Step by step solution

Step 1. Given information.

Consider the given question,

$\{\begin{cases} x^{3} - 2 x^{2} + y^{2} + 3 y - 4 = 0 . . . . . . (i) \\ x - 2 + \frac{y^{2} - y}{x^{2}} = 0 . . . . . . (i i) \end{cases}$

Multiply equation (ii) by $x^{2}$ ,

$x (x^{2}) - 2 (x^{2}) + \frac{y^{2} - y}{x^{2}} (x^{2}) = 0 (x^{2}) x^{3} - 2 x^{2} + y^{2} - y = 0 . . . . . . (i i i)$

Step 2. Substitute the value of y in equation (iii).

Subtract equations (i) and (iii),

$(x^{3} - 2 x^{2} + y^{2} + 3 y - 4) - (x^{3} - 2 x^{2} + y^{2} - y) = 0 4 y - 4 = 0 y - 1 = 0 y = 1$

Substitute the value of y in equation (iii),

$x^{3} - 2 x^{2} + {(1)}^{2} - (1) = 0 x^{2} (x - 2) = 0 x = 0, 2$

$x = 0$ cannot be considered as in equation (ii) x is in the denominator.

Therefore, the solution sets are $(2, 1)$ .

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Solve the system. Use any method you wish.
$\{\begin{cases} x^{3} - 2 x^{2} + y^{2} + 3 y - 4 = 0 \\ x - 2 + \frac{y^{2} - y}{x^{2}} = 0 \end{cases}$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Substitute the value of y in equation (iii).

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Solve the system. Use any method you wish.x3-2x2+y2+3y-4=0x-2+y2-yx2=0

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Substitute the value of y in equation (iii).

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Pure Maths

Decision Maths

Theoretical and Mathematical Physics

Statistics

Applied Mathematics

Study anywhere. Anytime. Across all devices.

Solve the system. Use any method you wish.
$\{\begin{cases} x^{3} - 2 x^{2} + y^{2} + 3 y - 4 = 0 \\ x - 2 + \frac{y^{2} - y}{x^{2}} = 0 \end{cases}$